# Irrationality or Rationality of p+q given that pq=1

Let $p,q$ are irrational numbers such that $pq=1$. Then what can we say about character of $p+q$? i.e, $p+q$ is rational or irrational? Probably, I believe it is irrational but could not prove.

Take $$p=\sqrt { 2 } ,q=\frac { 1 }{ \sqrt { 2 } } \\ \\ p=2-\sqrt { 3 } ,q=2+\sqrt { 3 } \\$$

• very nice !! ok I get it – Jamal Gadirov May 21 '17 at 19:10

The question then becomes, if $p$ is irrational, then is $p+\frac{1}{p}$ irrational?

In other words, you want to know if $$\frac{p^2+1}{p}=\frac{a}{b}$$ for some integers $a$ and $b$.

Let's just try a value for $\frac{a}{b}$. Let's try $\frac{a}{b}=3$. Therefore, we are trying to solve:

$$\frac{p^2+1}{p}=3.$$ This can turned into $$p^2-3p+1=0$$ and solved with the quadratic formula: In this case, $$p=\frac{3\pm\sqrt{9-4}}{2}=\frac{3}{2}\pm\frac{\sqrt{5}}{2},$$ which is irrational. Therefore, the answer to your question is that $p+q$ can be rational.

In general, we can find a $p$ that satisfies the conditions if $\frac{p^2+1}{p}\geq 2$. Most of the results from this are irrational. We can certainly replace $\frac{a}{b}$ with an irrational number greater than $2$ and get possibilities for $p$ and $\frac{1}{p}$ where their sum is irrational.

Hint: consider for example:

• $\;p=\sqrt{2}, q= 1 / \sqrt{2}$

• $\;p=3+2\sqrt{2}, q= 3 - 2 \sqrt{2}$

[ EDIT ] For more examples of irrationals with product $1$ and:

• irrational sum: let $p$ be a transcendental number (such as $\pi$ for example) and $q=1/p\,$, then $p+q$ will necessarily be transcendental, thus irrational; the same goes for $p$ being the sum between a transcendental number and an algebraic irrational (such as $\pi+\sqrt{2}$ for example);

• rational sum: let $p,q$ be the roots of the equation $x^2 - n \,x + 1 = 0\,$ for integer $n \gt 2$. Both $p,q$ are irrational by the rational root theorem, and $pq=1\,$, $p+q=n\,$ by Vieta's relations.

• ok that means both of them is possible – Jamal Gadirov May 21 '17 at 19:10

Let consider $S=p+q$, $P=pq=1$ and $p>q$

The roots of the polynom $P[X]=X²-SX+P=X²-SX+1$ are $p$ and $q$.

$discriminant = \delta = S²-4 > 0$ given that $P[X]$ has two roots and they are irrational numbers. Then, $p = \frac{S + \sqrt{S²-4}}{2}$ and $q = \frac{S - \sqrt{S²-4}}{2}$

As soon as $|S| > 2$ and $S²-4$ is not a perfect square, you can choose $S$ as you want, rational or irrational. $p$ and $q$ will be irrational and the sum will be what you have chosen.

There is another case, namely $p$ and $q=\frac{1}{p}$ are transcendental irrationals. In that case $p+q$ is necessarily transcendental, hence irrational, because $p+\frac{1}{p}=r$ with $r$ rational would imply $p^2-pr+1=0$ contradicting $p$ is transcendental.